Q.In J.J. Thomson's e/m experiment, a beam of electrons is replaced by that of muons (a particle with the same charge as an electron but mass 208 times that of an electron). The no-deflection condition is achieved only if

Step 1. The velocity-selector balance condition itself, $eE=evB\Rightarrow v=E/B$, does not involve mass at all - so simply keeping the beam undeflected in the crossed-field stage would not require changing $B$.

Step 2. However, the "no deflection condition" in this apparatus is set up together with the circular-motion (radius) stage of the measurement, where the magnetic force supplies the centripetal force, $evB=mv^2/R$, giving $R=mv/(eB)$. For the beam to trace out the same geometric path in the tube (same $R$) with the same velocity $v$ (fixed by $E/B$ at the balance point), a particle of mass $208m_e$ requires $B$ to be scaled up. …

Recognise that keeping the same beam geometry for a particle 208 times heavier requires …

• Assuming $B$ must scale by the full mass ratio (208) rather than its square root. …